Rs Aggarwal %282017%29 Solutions for Class 8 Math Chapter 11 Compound Interest are provided here with simple step-by-step explanations. These solutions for Compound Interest are extremely popular among Class 8 students for Math Compound Interest Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal %282017%29 Book of Class 8 Math Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal %282017%29 Solutions. All Rs Aggarwal %282017%29 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.
Page No 144:
Question 1:
Find the amount and the compound interest on Rs 2500 for 2 years at 10% per annum, compounded annually.
Answer:
Principal for the first year = Rs. 2500Interest for the first year = Rs. 2500×10×1 100= Rs. 250Amount at the end of the first year = Rs. (2500 +250 ) = Rs. 2750Principal for the second year = Rs. 2750Interest for the second year = Rs. 2750×10×1100= Rs. 275Amount at the end of the second year = Rs. (2750 + 275) = Rs. 3025∴ Compound interest = Rs. 3025 - 2500 = Rs. 525
Page No 144:
Question 2:
Find the amount and the compound interest on Rs 15625 for 3 years at 12% per annum, compounded annually.
Answer:
Principal for the first year = Rs. 15625Interest for the first year = Rs. 15625×12×1100= Rs. 1875Amount at the end of the first year = Rs. ( 15625 + 1875) = Rs. 17500Principal for the second year = Rs. 17500Interest for the second year = Rs. 17500×12×1100= Rs. 2100Amount at the end of the second year = Rs. (17500 + 2100 )= Rs. 19600Principal for the third year = Rs. 19600Interest for the third year = Rs. 19600×12×1100= Rs. 2352Amount at the end of the second year = Rs (19600 + 2352) = Rs. 21952∴ Compound interest = Rs. 21952 - 15625 = Rs. 6327
Page No 144:
Question 3:
Find the difference between the simple interest and the compound interest on Rs 5000 for 2 years at 9% per annum.
Answer:
Principal amount = Rs. 5000Simple interest = Rs. 5000×2×9100= Rs. 900The compound interest can be calculated as follows :Principal for the first year = Rs. 5000Interest for the first year = Rs. 5000×9×1100= Rs. 450Amount at the end of the first year = Rs. (5000 +450) = Rs. 5450Principal for the second year = Rs. 5450Interest for the second year = Rs. 5450×9×1100= Rs. 490.5Amount at the end of the second year = Rs. (5450 +490.5) = Rs. 5940.5∴ Compound interest = Rs. 5940.5 - 5000 = Rs. 940.5Now, difference between the simple interest and the compound interest = CI - SI =Rs. 940.5 - 900=Rs. 40.5
Page No 144:
Question 4:
Ratna obtained a loan of Rs 25000 from the Syndicate Bank to renovate her house. If the rate of interest is 8% per annum, what amount will she have to pay to the bank after 2 years to discharge her debt?
Answer:
Principal for the first year = Rs. 25000Interest for the first year = Rs. 25000×8×1100= Rs . 2000Amount at the end of the first year = Rs. (25000 +2000) = Rs. 27000Principal for the second year = Rs. 27000Interest for the second year = Rs. 27000×8×1100= Rs. 2160Amount at the end of the second year = Rs. (27000 +2160) = Rs. 29160Therefore, Ratna has to pay Rs. 29160 after 2 years to discharge her debt.
Page No 144:
Question 5:
Harpreet borrowed Rs 20000 from her friend at 12% per annum simple interest. She lent it to Alam at the same rate but compounded annually. Find her gain after 2 years.
Answer:
Principal amount = Rs. 20000Simple interest = Rs. 20000×2×12100= Rs. 4800The compound interest can be calculated as follows: Principal for the first year = Rs. 20000Interest for the first year = Rs. 20000×12×1100= Rs. 2400Now, amount at the end of the first year = Rs. (20000 + 2400)= Rs. 22400Principal for the second year = Rs. 22400Interest for the second year = Rs. 22400×12×1 100= Rs. 2688Now, amount at the end of the second year = Rs. (22400 + 2688) = Rs. 25088Hence, compound interest = Rs. 25088 - 20000 = Rs. 5088Now, CI - SI =Rs. 5088-4800= Rs. 288∴ The amount of money Harpreet will gain after two years is Rs 288.
Page No 144:
Question 6:
Manoj deposited a sum of Rs 64000 in a post office for 3 years, compounded annually at 712% per annum. What amount will he get on maturity?
Answer:
Principal for the first year = Rs. 64000Interest for the first year = Rs. 64000×15×1100×2= Rs. 4800Now , amount at the end of the first year = Rs.( 64000 +4800) = Rs. 68800 Principal for the second year = Rs. 68800Interest for the second year = Rs. 68800 ×15×1100×2= Rs. 5160Now, amount at the end of the second year = Rs. ( 68800 +5160) = Rs. 73960Principal for the third year = Rs. 73960Interest for the third year = Rs. 73960×15×1100×2= Rs. 5547Now , amount at the end of the third year = Rs. (73960 +5547) = Rs. 79507 ∴ Manoj will get an amount of Rs.79507 after 3 years.
Page No 144:
Question 7:
Divakaran deposited a sum of Rs 6250 in the Allahabad Bank for 1 year, compounded half-yearly at 8% per annum. Find the compound interest he gets.
Answer:
Principal amount = Rs. 6250Rate of interest = 8% per annum = 4% for half yearTime = 1 year = 2 half yearsPrincipal for the first half year = Rs. 6250Interest for the first half year = Rs. 6250 ×4×1100= Rs. 250Now, amount at the end of the first half year = Rs. (6250 +250) = Rs. 6500Principal for the second half year = Rs. 6500Interest for the second half year = Rs. 6500×4×1100= Rs. 260Now, amount at the end of the second half year = Rs (6500 + 260) = Rs. 6760∴ Compound interest = Rs 6760 - 6250 = Rs 510Hence, Divakaran gets a compound interest of Rs 510.
Page No 145:
Question 8:
Michael borrowed Rs 16000 from a finance company at 10% per annum, compounded half-yearly. What amount of money will discharge his debt after 112 years?
Answer:
Principal amount = Rs. 16000Rate of interest = 10% per annum = 5 % for half yearTime = 112 years = 3 half yearsPrincipal for the first half year = Rs. 16000Interest for the first half year = Rs. 16000×5×1 100= Rs. 800Now, amount at the end of the first half year = Rs. (16000 + 800) = Rs. 16800Principal for the second half year = Rs. 16800Interest for the second half year = Rs. 16800×5×1100= Rs. 840Now , amount at the end of the second half year = Rs. (16800 + 840) = Rs. 17640Principal for the third half year = Rs. 17640Interest for the third half year = Rs. 17640×5×1100= Rs. 882Now, amount at the end of the third half year = Rs. (17640 + 882) = Rs. 18522∴ The amount of money Michael has to pay the finance company after 112years is Rs 18522.
Page No 151:
Question 1:
By using the formula,
find the amount and compound interest on:
Rs 6000 for 2 years at 9% per annum compounded annually.
Answer:
Principal amount, P = Rs 6000Rate of interest , R= 9% per annumTime, n =2 years.The formula for the amount including the compound interest is given below:A = Rs. P1+R100n ⇒A = Rs. 6000 1+91002⇒A = Rs. 6000 100+9100 2⇒A = Rs. 6000 1091002⇒A = Rs. 6000 1.09×1 .092⇒A = Rs. 7128.6i.e., the amount including the compound interest is Rs 7128.6.∴ Compound interest=Rs 7128.6 -6000=Rs 1128.6
Page No 151:
Question 2:
By using the formula, find the amount and compound interest on:
Rs 10000 for 2 years at 11% per annum compounded annually.
Answer:
Principal amount, P=Rs. 10000Rate of interest, R=11% per annum.Time, n=2 years.The formula for the amount including the compound interest is given below:A = Rs. P1+R100n ⇒A = Rs. 10000 1+111002⇒A = Rs. 10000 100 +111002⇒A = Rs.10000 1111002⇒A = Rs.10000 1 .11×1.112⇒A = Rs. 12321i.e., the amount including the compound interest is Rs 12321.∴ Compound interest =Rs. 12321 -10000 = Rs. 2321
Page No 151:
Question 3:
By using the formula, find the amount and compound interest on:
Rs 31250 for 3 years at 8% per annum compounded annually.
Answer:
Principal amount, P = Rs. 31250Rate of interest, R= 8% per annum.Time, n =3 years.The formula for the amount including the compound interest is given below:A = Rs. P 1+R100 n ⇒A = Rs. 31250 1+81003⇒A = Rs. 31250 100+81003⇒A = Rs.31250 1081003⇒A = Rs.31250 1.08×1.08×1.083⇒A = Rs. 39366i.e., the amount including the compound interest is Rs 39366.∴ Compound interest=Rs. 39366 -31250=Rs. 8116
Page No 151:
Question 4:
By using the formula, find the amount and compound interest on:
Rs 10240 for 3 years at 1212% per annum compounded annually.
Answer:
Principal amount, P = Rs. 10240Rate of interest, R= 1212% p.a. Time, n =3 yearsThe formula for the amount including the compound interest is given below: A = Rs. P1+R100n ⇒A = Rs. 10240 1+25100×2 3⇒A = Rs. 10240 1+252003⇒A = Rs. 10240 1+183⇒A = Rs. 10240 8+183⇒A = Rs. 10240 983⇒A = Rs. 10240 1.125×1.125×1.1253 ⇒A = Rs. 14580i.e., the amount including the compound interest is Rs 14580.∴ Compound interest=Rs 14580 -10240 = Rs. 4340
Page No 151:
Question 5:
By using
the formula, find the amount and compound interest on:
Rs 10240 for 3 years at 1212% per annum compounded annually.
Answer:
Principal amount, P = Rs 62500 Rate of interest, R= 12% p.a.Time, n =2 years 6 months=52=212 yearsThe formula for the amount including the compound interest is given below:A = Rs. P1+R100n ⇒A = Rs. 62500 1+121002×1+12×12100⇒A = Rs. 62500 1+12 1002×1+6100⇒A = Rs. 62500 ×1.12×1.12×1.06⇒A = Rs. 83104i.e., the amount including the compound interest is Rs 83104.∴ Compound interest=Rs. 83104-62500 = Rs. 20604
Page No 151:
Question 6:
By using the formula, find the amount and compound interest on:
Rs 9000 for 2 years 4 months at
10% per annum compounded annually.
Answer:
Principal amount, P=Rs. 9000Rate of interest, R= 10% p .a.Time, n =2 years 4 months = 213years = 73 years The formula for the amount including the compound interest is given below:A = Rs. P ×1+R100n = Rs. 9000×1+101002×1+13×10100= Rs. 9000×1.10×1.10×1.033= Rs. 11252.9≈11253i.e., the amount including the compound interest is Rs 11253. ∴ Compound interest=Rs. 11253 -9000 = Rs. 2253
Page No 151:
Question 7:
Find the amount of Rs 8000 for 2 years compounded annually and the rates being 9% per annum during the first year and 10% per annum during the second year.
Answer:
Principal amount, P = Rs. 8000Rate of interest for the first year , p= 9% p.a.Rate of interest for the second year, q= 10% p.a.Time, n =2 years.Formula for the amount including the compound interest for the first year:A = Rs. P×1+p100×1+q 100 = Rs. 8000×1+9100×1+10100 = Rs. 8000×109100×110100 = Rs . 8000× 1.09×1.1= Rs. 9592i.e., the amount including the compound interest for first year is Rs 9592.
Page No 151:
Question 8:
Anand obtained a loan of Rs 125000 from the Allahabad Bank for buying computers. The bank charges compound interest at 8% per annum, compounded annually. What amount wil he have to pay after 3 years to clear the debt?
Answer:
Principal amount, P=Rs. 125000 Rate of interest, R= 8% p.a.Time, n =3 yearsThe amount including the compound interest is calculated using the formula,A = Rs. P1+ R100n = Rs. 125000 1+81003 = Rs. 125000 100+81003 = Rs. 125000 1081003= Rs. 125000 1.08 3= Rs. 125000 1.08×1.08×1.08= Rs. 157464∴ Anand has to pay Rs 157464 after 3 years to clear the debt.
Page No 151:
Question 9:
Three years ago, Beeru purchased a buffalo from Surjeet for Rs 11000. What payment will discharge his debt now, the rate of interest being 10% per annum, compounded annually?
Answer:
Principal amount, P = Rs. 11000Rate of interest, R= 10% p.a.Time, n =3 yearsThe amount including the compound interest is calculated using the formula,A = Rs. P 1+R100n = Rs. 11000 1+101003 = Rs. 11000 100+101003 = Rs.11000 1101003 = Rs.11000 1.13 = Rs. 11000 1.1×1.1×1.1 = Rs. 14641Therefore, Beeru has to pay Rs 14641 to clear the debt.
Page No 151:
Question 10:
Shubhalaxmi took a loan of Rs 18000 from surya Finance to purchase a TV set. If the company charges compound interest at 12% per annum during the first year and 1212 % per annum during the second year, how much will she have to pay after 2 years?
Answer:
Principal amount, P = Rs. 18000Rate of interest for the first year, p= 12% p.a.Rate of interest for the second year , q= 1212% p.a.Time, n =2 yearsThe formula for the amount including the compound interest for the first year is given below: A = P×1+p100×1+q100 = Rs. 18000×1+12100×1+25100×2= Rs. 18000 ×100+12100×1+25200= Rs. 18000×100+12100×1+18 = Rs. 18000× 100+12100×8+18 = Rs. 18000×112 100×98= Rs. 18000× 1.12×1.125 = Rs. 22680∴ Shubhalaxmi has to pay Rs 22680 to the finance company after 2 years.
Page No 151:
Question 11:
Neha borrowed Rs 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?
Answer:
Principal amount, P = Rs. 24000Rate of interest, R= 10% p.a.Time, n =2 years 3 months = 214 yearsThe formula for the amount including the compound interest is given below:A = P ×1+R100n× 1+14R100= Rs. 24000 ×1+101002×1+ 14×10100= Rs. 24000× 100+101002×100+2.5100 = Rs. 24000× 1101002×100+2.5100=Rs. 24000×1.1× 1.1×1.025= Rs. 24000×1.250= Rs. 29766Therefore, Neha should pay Rs 29766 to the bank after 2 years 3 months.
Page No 151:
Question 12:
Abhay borrowed Rs 16000 at 712 % per annum simple interest. On the same day, he lent it to Gurmeet at the same rate but compounded annually. What does he gain at the end of 2 years?
Answer:
Principal amount, P = Rs 16000Rate of interest, R= 152% p.a.Time, n =2 yearsNow, simple interest = Rs 16000×2×15100×2= Rs. 2400Amount including the simple interest = Rs 16000+2400= Rs 18400The formula for the amount including the compound interest is given below:A = P 1+R100n = Rs. 16000 1+15100×22 = Rs. 16000 1+152002 = Rs.16000 1+3402= Rs.16000 40+3402 = Rs. 16000 43402 = Rs. 16000 1.075×1.075i. e., the amount including the compound interest is Rs 18490.Now, CI - SI= Rs. 18490 -18400 = Rs. 90Therefore, Abhay gains Rs. 90 as profit at the end of 2 years.
Page No 151:
Question 13:
The simple interest on a sum of money for 2 years at 8% per annum is Rs 2400. What will be the compound interest on that sum at the same rate and for the same period?
Answer:
Simple interest (SI) = Rs. 2400Rate of interest, R = 8% Time, n = 2 yearsThe principal can be calculated using the formula:Sum = 100×SIR×T⇒Sum = Rs. 100×24008×2= Rs. 15000i.e., the principal is Rs. 15000.The amount including the compound interest is calculated using the formula given below:A = P 1+R100 n = Rs. 15000 1+81002 = Rs. 15000 100+8 1002 = Rs. 15000 1081002= Rs. 15000 1.08×1.08 = Rs. 17496i.e., the amount including the compound interest is Rs. 17496.∴ Compound interest (CI) = Rs. 17496 - 15000=Rs. 2496
Page No 151:
Question 14:
The difference between the compound interest and the simple interest on a certain sum for 2 years at 6% per annum is Rs 90. Find the sum.
Answer:
Let Rs P be the sum. Then SI = P×2×6100= Rs. 12P100= Rs.3P25Also, CI = P ×1+61002-P= Rs. P×100+61002-P = Rs. P×53502-P= Rs. 2809P2500-P = Rs. 2809P-2500P2500= Rs.309P2500Now, CI - SI=Rs. 309P2500- 3P25= Rs. 309P-300P2500 = Rs. 9P2500Now, Rs. 90 = 9P2500⇒P = 90 ×25009= Rs. 25000Hence, the required sum is Rs. 25000.
Page No 151:
Question 15:
The difference between the compound interest and the simple interest on a certain sum for 3 years at 10% per annum is Rs 93. Find the sum.
Answer:
Let P be the sum. Then SI = Rs P×3×10100= Rs 30P100= Rs 3P10Also, CI = Rs. P×1+101003-P = Rs. P×100+10 1003-P = Rs. P×11103-P = Rs. 1331P1000-P= Rs. 1331P-1000P1000= Rs.331P1000 Now, CI - SI=Rs 331P1000- 3P10= Rs 331P-300P1000= Rs 31P1000Now, Rs. 93 = 31P1000⇒P = 93×100031= Rs. 3000Hence, the required sum is Rs. 3000.
Page No 152:
Question 16:
A sum of money amounts to Rs 10240 in 2 years at 623% per annum, compounded annually. Find the sum.
Answer:
Let P be the sum. Rate of interest, R = 623% = 203%Time, n = 2 yearsNow, A= P×1+20100×32= Rs. P×1+203002= Rs. P×300+203002= Rs. P× 3203002 = Rs. P× 1615×1615= Rs . 256P225⇒Rs. 10240 = Rs. 256P225⇒Rs. 10240×225256 = P∴ P = Rs. 9000Hence, the required sum is Rs . 9000
Page No 152:
Question 17:
What sum of money will amount to Rs 21296 in 3 years at 10% per annum, compounded annually?
Answer:
Let P be the sum.Rate of interest, R =10%Time, n = 3 yearsNow, A= P×1+10 1003= Rs. P×100+101003= Rs. P× 110100 3 = Rs. P× 1110×1110×1110= Rs. 1331P 1000However, amount = Rs. 21296Now, Rs. 21296 = Rs. 1331P1000 ⇒Rs. 21296×10001331= P∴ P = Rs. 16000Hence, the req uired sum is Rs. 16000.
Page No 152:
Question 18:
At what rate per cent per annum will Rs 4000 amount to Rs 4410 in 2 years when compounded annually?
Answer:
Let R% p.a. be the required rate. A = 4410P = 4000n = 2 yearsNow, A = P 1+R100n⇒4410 = 4000 1 +R1002⇒44104000= 1+R1002⇒441400= 1+R100 2⇒21202= 1+R1002⇒2120-1= R100⇒21-2020= R100⇒120=R100⇒R = 1×10020= 5Hence, the required rate is 5% p.a.
Page No 152:
Question 19:
At what rate per cent per annum will Rs 640 amount to Rs 774.40 in 2 years when compounded annually?
Answer:
Let the required rate be R% p.a.A = 774.40P = 640 n = 2 yearsNow, A = P 1+R100n⇒774.40 = 640 1+R1002⇒774 .40640 = 1+R1002⇒1.21= 1+R100 2⇒1.12= 1+R1002⇒1.1-1= R100⇒ 0.1= R100⇒R = 0.1×100= 10Hence, the required rate is 10% p.a.
Page No 152:
Question 20:
In how many years will Rs 1800 amount to Rs 2178 at 10% per annum when compounded annually?
Answer:
Let the required time be n years.Rate of interest, R = 10%Principal amount, P = Rs. 1800Amount with compound interest, A = Rs. 2178Now, A = P× 1+R100n= Rs. 1800×1+10100n = Rs. 1800× 100+10100n=Rs. 1800×110100n = Rs. 1800×11 10nHowever, amount =Rs. 2178Now, Rs. 2178 = Rs. 1800×11 10n⇒21781800 = 1110n⇒121100 = 1110n ⇒11102 = 1110n⇒n = 2∴ Time, n = 2 years
Page No 152:
Question 21:
In how many years will Rs 6250 amount to Rs 7290 at 8% per annum, compounded annually?
Answer:
Let the required time be n years. Rate of interest, R = 8%Principal amount, P = Rs. 6250Amount with compound interest, A = Rs. 7290Then, A = P× 1+R100n⇒A= Rs. 6250×1+8100n= Rs. 6250× 100+8100n=Rs. 6250×108100n =Rs. 6250×2725nHowever, amount = Rs. 7290Now, Rs. 7290 = Rs. 6250×2725 n⇒72906250 = 2725n⇒729625 = 2725n⇒ 27252 = 2725n⇒n = 2∴ Time, n = 2 years
Page No 152:
Question 22:
The population of a town is 125000. It is increasing at the rate of 2% per annum. What will be its population after 3 years?
Answer:
Population of the town, P= 125000Rate of increase, R= 2%Time, n = 3 yearsThen the population of the town after 3 years is given byPopulation = P×1+R1003= 125000×1+21003= 125000×100+2100 3 = 125000×1021003 = 125000×51503= 125000 × 5150×5150×5150= 51×51×51 = 132651Therefore, the population of the town after three years is 132651.
Page No 152:
Question 23:
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 5%, 4% and 3% respectively, what is its present population?
Answer:
Let the population of the town be 50000.Rate of increase for the first year, p = 5%Rate of increase for the second year, q = 4% Rate of increase for the third year, r = 3%Time = 3 yearsNow, present population= P ×1+p100×1+q100×1+r100 =50000 ×1+5100×1+4100×1+3100=50000 × 100+5100×100+4100×100+3100=50000 ×105100× 104100×103100=50000 ×2120×2625×103100 =21×26×103=56238Therefore, the present population of the town is 56238.
Page No 152:
Question 24:
The population of a city was 120000 in the year 2009. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. what is its population in the year 2011?
Answer:
Population of the city in 2009, P= 120000Rate of increase, R= 6%Time, n = 3 yearsThen the population of the city in the year 2010 is given byPopulation = P×1+R100n= 120000×1+61001 = 120000×100 +6100 = 120000×106100 = 120000×5350 = 2400 ×53 = 127200Therefore, the population of the city in 2010 is 127200. Again, population of the city in 2010, P=127200Rate of decrease, R= 5%Then the population of the city in the year 2011 is given byPopulation = P×1-R100n= 127200×1-51001= 127200×100-5100 = 127200×95100 = 127200×1920= 6360 ×19 = 120840Therefore, the population of the city in 2011 is 120840.
Page No 152:
Question 25:
The count of bacteria in a certain experiment was increasing at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 500000.
Answer:
Initial count of bacteria, P= 500000Rate of increase, R= 2%Time, n = 2 hoursThen the count of bacteria at the end of 2 hours is given byCount of bacteria = P×1+R100n =500000×1+21002=500000×100+21002=500000×1021002 =500000×51502=500000×5150×5150=200×51×51=520200Therefore, the count of bacteria at the end of 2 hours is 520200.
Page No 152:
Question 26:
The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if the count was initially 20000.
Answer:
Initial count of bacteria, P= 20000Rate of increase, R= 10%Time, n = 3 hoursThen the count of bacteria at the end of the first hour is given byCount of bacteria = P×1+10100n=20000×1+101001=20000×100+10100 =20000×110100=20000×1110=2000×11=22000Therefore, the count of bacteria at the end of the first hour is 22000.The count of bacteria at the end of the second hour is given byCount of bacteria = P×1-10100 n=22000×1-101001=22000×100-10100=22000×90100 =22000×910=2200×9=19800Therefore, the count of bacteria at the end of the second hour is 19800.Then the count of bacteria at the end of the third hour is is given byCount of bacteria = P×1+10 100n=19800×1+101001=19800×100+10100=19800× 110100=19800×1110=1980×11=21780Therefore, the count of bacteria at the end of the first 3 hours is 21780.
Page No 152:
Question 27:
A machine is purchased for Rs 625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years?
Answer:
Initial value of the machine, P= Rs 625000Rate of depreciation, R= 8%Time, n = 2 yearsThen the value of the machine after two years is given byValue = P×1-R100n=Rs 625000×1- 81002=Rs 625000×100-81002=Rs 625000×921002= Rs 625000×23252=Rs 625000×2325×2325=Rs 1000×23×23 =Rs 529000Therefore, the value of the machine after two years will be Rs. 529000 .
Page No 152:
Question 28:
A scooter is bought at Rs 56000. Its value depreciates at the rate of 10% per annum. What will be its value after 3 years?
Answer:
Initial value of the scooter , P= Rs 56000Rate of depreciation, R= 10%Time, n = 3 yearsThen the value of the scooter after three years is given byValue = P×1- R100n=Rs. 56000×1-101003=Rs. 56000×100-101003 =Rs. 56000×901003=Rs. 56000×9103=Rs. 56000× 910×910×910=Rs. 56×9×9×9=Rs. 40824Therefore , the value of the scooter after three years will be Rs. 40824.
Page No 152:
Question 29:
A car is purchased for Rs 348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years?
Answer:
Initial value of the car, P= Rs 348000Rate of depreciation for the first year, p= 10%Rate of depreciation for the second year, q= 20%Time, n = 2 years.Then the value of the car after two years is given byValue = P×1-p100×1-q100=Rs. 348000×1-10100×1-20100=Rs. 348000×100-10100×100-20100=Rs. 348000×90100 ×80100=Rs. 348000×910×810=Rs. 3480×9×8=Rs. 250560∴ The value of the car after two years is Rs 250560.
Page No 152:
Question 30:
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 291600, for how much was it purchased?
Answer:
Let the initial value of the machine, P be Rs x.Rate of depreciation, R= 10%Time, n = 3 yearsThe present value of the machine is Rs 291600.Then the initial value of the machine is given byValue = P×1-R100n=Rs. x×1 -101003=Rs. x×100-101003=Rs. x×901003=Rs. x×9103∴ Present value of the machine = Rs 291600Now, Rs 291600 = Rs x×910×910×910⇒x = Rs 291600×10×10×109×9×9⇒x =Rs 291600000729⇒x = Rs 400000∴ The initial value of the machine is Rs 400000.
Page No 154:
Question 1:
Find the amount and the compound interest on Rs 8000 for 1 year at 10% per annum, compounded half-yearly.
Answer:
Principal, P = Rs. 8000Time, n = 1 year = 2 half yearsRate of interest per annum = 10%Rate of interest for half year, R = 10%2= 5%The amount with the compound interest is given byAmount = Rs. P×1+R1002= Rs. 8000×1+51002 = Rs. 8000×1051002 = Rs. 8000×21202 = Rs. 8000×21 20×2120 = Rs. 20×21×21= Rs. 8820∴ Compound interest = amount - principal = Rs. (8820 -8000) = Rs. 820
Page No 154:
Question 2:
Find the amount and the compound interest on Rs 31250 for 112 years at 8% per annum, compounded half-yearly.
Answer:
Principal, P = Rs. 31250Annual rate of interest, R = 8% Rate of interest for a half year = 12×8%= 4%Time, n = 112 years = 3 half yearsThen the amount with the compound interest is given by A = P ×1+R100 n = 31250×1+41003 = 31250×1+1253 = 31250×25+1253 = 31250×26253 = 31250×2625×2625×2625 = Rs 2×17576 = Rs 35152 Therefore, compound interest = amount - principal= Rs (35152- 31250) = Rs 3902
Page No 154:
Question 3:
Find the amount and the compound interest on Rs 12800 for 1 year at 712% per annum, compounded semi-annually.
Answer:
Principal, P = Rs. 12800Annual rate of interest, R = 152% Rate of interest for a half year = 12 152%= 154%Time, n = 1 year = 2 half yearsThen the amount with the compound interest is given by A = P ×1+R100n = 12800×1+1541002= 12800×1+15100×42 = 12800×400+ 154002 = 12800×4154002 = 12800×8380×8380 = 2×83×83 =Rs 13778Therefore, compound interest = amount - principal=Rs ( 13778-12800) = Rs 978
Page No 154:
Question 4:
Find the amount and the compound interest on Rs 160000 for 2 years at 10% per annum, compounded half-yearly.
Answer:
Principal, P = Rs. 160000Annual rate of interest, R = 10% Rate of interest for a half year = 102%= 5%Time, n = 2 years = 4 half yearsThen the amount with the compound interest is given by A = P ×1+R100n= 160000×1+5100 4 = 160000×100+51004 = 160000×1051004 = 160000×2120×2120×2120×2120= 21×21×21×21 = Rs 194481Therefore, compound interest = amount - principal=Rs (194481-160000) = Rs 34481
Page No 154:
Question 5:
Swati borrowed Rs 40960 from a bank to buy a piece of land. If the bank charges 1212% per annum, compounded half-yearly, what amount will she have to pay after 112 years? Also, find the interest paid by her.
Answer:
Principal, P = Rs. 40960Annual rate of interest, R = 25 2% Rate of interest for half year = 254%Time, n = 112 years = 3 half yearsThen the amount with the compound interest is given by A = P ×1+R100n= 40960×1+25100×43 = 40960×400+254003 = 40960×4254003 = 40960×1716× 1716×1716 = 10×17×17×17 = Rs 49130Therefore, compound interest = amount - principal=Rs (49130- 40960) = Rs 8170Therefore, Swati has to pay Rs. 49130, which includes an interest of Rs. 8170, to the bank after 112 years.
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Question 6:
Mohd. Aslam purchased a house from Avas Vikas Parishad on credit. If the cost of the house is Rs 125000 and the Parishad charges interest at 12% per annum compounded half-yearly, find the interest paid by Aslam after a year and a half.
Answer:
Let the principal amount be P=Rs. 125000.Annual rate of interest, R = 12% Rate of interest for a half year = 6%Time, n = 112 years = 3 half yearsThen the amount with the compound interest is given by A = P ×1+R100n= Rs. 125000×1+61003 = Rs. 125000×100+61003 = Rs. 125000×1061003 = Rs. 125000×5350×5350×5350= Rs. 53×53×53 = Rs. 148877Now, CI = A - P=Rs. (148877-125000) = Rs. 23877 Therefore, Aslam has to pay an interest of Rs. 23877 to the bank after 112 years.
Page No 154:
Question 7:
Sheela deposited Rs 20000 in a bank, where the interest is credited half-yearly. If the rate of interest paid by the bank is 6% per annum, what amount will she get after 1 year?
Answer:
Let the principal amount be P= Rs. 20000.Annual rate of interest, R = 6% Rate of interest for half year = 3%Time, n=1 year = 2 half yearsThen the amount with the compound interest is given by A = P ×1+R100n= Rs. 20000×1+31002 = Rs. 20000×100+31002 = Rs. 20000×1031002= Rs. 20000×103100×103100 = Rs. 2 ×103×103= Rs. 21218Therefore, Sheela gets Rs. 21218 after 1 year.
Page No 154:
Question 8:
Neeraj lent Rs 65536 for 2 years at 1212% per annum, compounded annually. How much more could he earn if the interest were compounded half-yearly?
Answer:
Let the principal amount be P= Rs. 65536.Annual rate of interest, R = 252% Rate of interest for a half year = 254%Time, n = 2 years = 4 half yearsThen the amount with the compound interest is given by A = P × 1+R100n= Rs. 65536×1+25100×44 = Rs. 65536 ×400+254004 = Rs. 65536×4254004= Rs. 65536×17 164= Rs. 65536×1716×1716×1716×1716 = Rs. 17×17×17×17 = Rs. 83521Now, CI = A - P = Rs. (83521 -65536) = Rs. 17985Therefore, interest earned when compounded half yearly = Rs. 17985Amount when the interest is compounded yearly is given by A= P ×1+R100n=Rs. 65536×1+25100 ×22= Rs. 65536×200+252002 = Rs. 65536×225200 2= Rs. 65536×982= Rs. 65536×98×98 =Rs. 82944Therefore, CI = A- P =Rs. (82944-65536)=Rs. 17408∴ Difference between the interests compounded half yearly and yearly =Rs. (17985-17408)=Rs. 577
Page No 154:
Question 9:
Sudershan deposited Rs 32000 in a bank, where the interest is credited quarterly. If the rate of interest be 5% per annum, what amount will he receive after 6 months?
Answer:
Let the principal amount be P=Rs 32000. Annual rate of interest, R = 5% Rate of interest for a quarter year = 54 %Time, n=6 months= 2 quarter yearsThen the amount with the compound interest is given by A = Rs. P ×1+R100n = Rs. 32000×1+5100×42= Rs. 32000×400+54002 = Rs. 32000×4054002 = Rs. 32000×81802 = Rs. 32000×8180×8180 = Rs. 5×81×81 = Rs. 32805Therefore, Sudershan will receive an amount of Rs. 32805 after 6 months.
Page No 154:
Question 10:
Arun took a loan of Rs 390625 from Kuber Finance. If the company charges interest at 16% per annum, compounded quarterly, what amount will discharge his debt after one year?
Answer:
Let the principal amount be P = Rs 390625.Annual rate of interest, R = 16% Rate of interest for a quarter year =164%=4%Time, n=1 year = 4 quarter yearsThen the amount with the compound interest is given by A = Rs. P ×1+R100n = Rs. 390625×1+41004 = Rs. 390625×100+41004 = Rs. 390625×1041004 = Rs. 390625×26254 = Rs. 390625×2625×2625×2625×2625= Rs. 26×26×26×26 = Rs. 456976Therefore, Arun has to pay Rs 456976 after 1 year.
Page No 154:
Question 1:
Tick (✓) the correct answer:
The compound interest on Rs 5000 at 8% per annum for 2 years, compounded annually, is
(a) Rs 800
(b) Rs 825
(c) Rs 832
(d) Rs 850
Answer:
(c) Rs. 832
A = P ×1+ R100n = Rs. 5000×1+81002 = Rs. 5000×108100 2 = Rs. 5000×27252 = Rs. 5000×2725×2725 = Rs. 8×27×27 = Rs. 5832∴ Interest = amount - principal = Rs (5832-5000)= Rs 832
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Question 2:
Tick (✓) the correct answer:
The compound interest on Rs 10000 at 10% per annum for 3 years, compounded annually, is
(a) Rs 1331
(b) Rs 3310
(c) Rs 3130
(d) Rs 13310
Answer:
(b) Rs. 3310
A = P ×1+R100n= Rs. 10000×1+101003 = Rs. 10000×1101003 = Rs. 10000×11103= Rs . 10000×1110×1110×1110 = Rs. 10×11×11×11 = Rs. 13310∴ Compound interest =amount - principal= Rs (13310 - 10000) = Rs 3310
Page No 154:
Question 3:
Tick (✓) the correct answer:
The compound interest on Rs 10000 at 12% per annum for 112 years, compounded annually, is
(a) Rs 1872
(b) Rs 1720
(c) Rs 1910.16
(d) Rs 1782
Answer:
(a) Rs 1872
Here, A = P ×1+R1001×1+12R100 = Rs 10000×1+12100×1+12×12100= Rs 10000×100 +12100×100+6100 = Rs 10000×112100×106100= Rs 10000×2825×5350 = Rs 8×28×53 = Rs 11872∴ Compound interest =amount -principal= Rs ( 11872-10000)= Rs 1872
Page No 155:
Question 4:
Tick (✓) the correct answer:
The compound interest on Rs 4000
at 10% per annum for 2 years 3 months, compounded annually, is
(a) Rs 916
(b) Rs 900
(c) Rs 961
(d) Rs 896
Answer:
(c) Rs 961
Here, A = P ×1+R100 2×1+14R100= Rs. 4000×1+101002×1+14×10100 = Rs. 4000×100+101002×40+140 = Rs. 4000×1101002×4140 = Rs. 4000×1110×11 10×4140 = Rs. 11×11×41 = Rs. 4961∴ Compound interest = amount -principal= Rs ( 4961- 4000)= Rs 961
Page No 155:
Question 5:
Tick (✓) the correct answer:
A sum of Rs 25000 was given as loan on
compound interest for 3 years compounded annually at 5% per annum during the first year, 6% per annum during the second year and 8% per annum during the third year. The compound interest is
(a) Rs 5035
(b) Rs 5051
(c) Rs 5072
(d) Rs 5150
Answer:
(b) Rs. 5051
Here, A = Rs. P ×1+p100×1+q100×1+r100 = Rs. 25000×1 +5100×1+6100×1+8100 = Rs. 25000×105100×106100 ×108100= Rs. 25000×2120×5350×2725 = Rs. 21×53×27= Rs. 30051∴ Compound interest = amount -principal= Rs. (30051 - 25000)= Rs. 5051
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Question 6:
Tick (✓) the correct answer:
The compound interest on Rs 6250 at 8% per annum for 1 year, compounded half yearly, is
(a) Rs 500
(b) Rs 510
(c) Rs 550
(d) Rs 512.50
Answer:
(b) Rs. 510
Rate of interest compounded half yearly = 82%=4%Time = 1 year= 2 half years Now, A = P ×1+R100n = Rs. 6250×1+ 41002= Rs. 6250×1041002= Rs. 6250×2625×2625 = Rs. 10×26×26 = Rs. 6760∴ Compound interest =amount- principal= Rs. (6760-6250)= Rs. 510
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Question 7:
Tick (✓) the correct answer:
The compound interest on Rs 40000 at 6% per annum for 6 months, compounded quarterly, is
(a) Rs 1209
(b) Rs 1902
(c) Rs 1200
(d) Rs 1306
Answer:
(a) Rs.1209
Time = 6 months = 2 quater yearsRate compounded quarter yearly =6 4%=32%Now, A = P ×1+R100n = Rs. 40000× 1+3100×22= Rs. 40000×2032002 = Rs. 40000×203 200×203200 = Rs. 203×203 = Rs. 41209∴ Compound interest = amount-principal= Rs. 41209-Rs. 40000= Rs.1209
Page No 155:
Question 8:
Tick (✓) the correct answer:
The present population of a town is 24000. If it increases at the rate of
5% per annum, what will be its population after 2 years?
(a) 26400
(b) 26460
(c) 24460
(d) 26640
Answer:
(b) 26460
Here, A = P ×1+R100n = Rs. 24000×1+51002 = Rs. 24000×1051002= Rs. 24000×2120×2120 = Rs. 60×21×21 = Rs. 26460
Page No 155:
Question 9:
Tick (✓) the correct answer:
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago for Rs 60000. What is the present value of the machine?
(a) Rs 53640
(b) Rs 51680
(c) Rs 43740
(d) Rs 43470
Answer:
(c) Rs. 43740
Here, A = Rs. P ×1-R100n = Rs. 60000×1-10100 3 = Rs. 60000×901003 = Rs. 60000×910×910 ×910 = Rs. 60×9×9×9= Rs. 43740
Page No 155:
Question 10:
Tick
(✓) the correct answer:
The value of a machine depreciates at the rate of 20% per annum. It was purchased 2 years ago. If its present value is Rs 40000, for how much was it purchased?
(a) Rs 56000
(b) Rs 62500
(c) Rs 65200
(d) Rs 56500
Answer:
(b) Rs. 62500
Here, A = P ×1-R100n = P×1-201002 = P×801002 = P×45×45⇒40000 = 16P25∴ P = 40000×2516= Rs 62500
Page No 155:
Question 11:
Tick (✓) the correct answer:
The annual rate of growth in population of a town is 10%. If its present population is 33275, what was it 3 years ago?
(a) 25000
(b) 27500
(c) 30000
(d) 26000
Answer:
(a) 25000
Let P be the popultion 3 years ago.Now, present population =33275⇒33275= P×1+101003⇒33275 = P×1101003⇒33275 = P×1110×1110×1110⇒ 33275 = 1331P1000∴ P = 33275×10001331= 25000
Page No 155:
Question 12:
Tick (✓) the correct answer:
If the simple interest on a sum of money at 5% per annum for 3 years is Rs 1200 then the compound interest on the same sum for the same period at the same rate will be
(a) Rs 1225
(b) Rs
1236
(c) Rs 1248
(d) Rs 1261
Answer:
(d) Rs 1261
Here, SI= P×5×3100⇒1200 =P×5×3100 ⇒P = 1200×1005×3 =Rs 8000Amount at the end of 3 years = Rs 8000×1+51003 = Rs 8000×1051003 = Rs 8000 ×2120×2120×2120 = Rs 21×21×21 = Rs 9261 ∴ CI = A-P = Rs ( 9261-8000)= Rs 1261
Page No 155:
Question 13:
Tick (✓) the correct answer:
If the compound interest
on a sum for 2 years at 1212% per annum is Rs 510, the simple interest on the same sum at the same rate for the same period of time is
(a) Rs 400
(b) Rs 450
(c) Rs 460
(d) Rs 480
Answer:
(d) Rs 480
We have: 510 =P×1+25100×22-P⇒510=⇒ P×8+182 -P⇒510 = P×98×98-P⇒510= 81P64-P⇒510 =81P-64P64 ⇒510 = 17 P64∴ P = 510×6417=Rs 1920Now, SI =P×R×T100 =Rs 1920×2×25100×2= Rs 480
Page No 155:
Question 14:
Tick (✓) the correct answer:
The sum that amounts to Rs 4913 in 3 years at 614% per annum compounded annually, is
(a) Rs 3096
(b) Rs 4076
(c) Rs 4085
(d) Rs 4096
Answer:
(d) Rs 4096
We have Rs 4913 = P ×1+25100×43⇒Rs 4913 = P×16+1163⇒Rs 4913 =P×1716×1716×1716⇒Rs 4913 = 4913P4096⇒P = Rs 4913×40964913=Rs 4096
Page No 155:
Question 15:
Tick (✓) the correct answer:
At what rate per cent per
annum will a sum of Rs 7500 amount to Rs 8427 in 2 years, compounded annually?
(a) 4%
(b) 5%
(c) 6%
(d) 8%
Answer:
(c) 6%
Here, A = P ×1+R100= Rs. 7500×1+R1002= Rs. 7500×1+R1002However, amount =Rs. 8427Now, Rs. 8427== Rs. 7500×1+R1002⇒Rs. 8427 Rs. 7500 = 1+R1002⇒53502= 1+R1002⇒1+R100 = 5350⇒R100= 5350-1⇒R100= 53-5050=350∴ R = 30050=6%
Page No 157:
Question 1:
Find the amount and the compound interest on Rs 3000 for 2 years at 10% per annum.
Answer:
Here, A = P ×1+R100n = Rs. 3000 × 1+101002= Rs. 3000 ×1101002 = Rs. 3000 × 1110×1110 = Rs. 30×11×11 = Rs. 3630∴ CI = A-P = Rs. (3630-3000) = Rs. 630Hence, the amount is Rs. 3630 and the CI is Rs. 630.
Page No 157:
Question 2:
Find the amount of Rs 10000 after 2 years compounded annually; the rate of interest being 10% per annum during the first year and 12% per annum during the second year. Also, find the compound interest.
Answer:
Here, A = P ×1+p100×1+r100 = Rs. 10000 ×1+10100×1+12100 = Rs. 10000 ×110100 ×112100 = Rs. 10000 ×1110×2825 = Rs. 40 ×11×28 = Rs. 12320∴ CI =A-P = Rs (12320-10000) = Rs. 2320Hence, the amount is Rs 12320 and the CI is Rs 2320.
Page No 157:
Question 3:
Find the amount and the compound interest on Rs 6000 for 1 year at 10% per annum compounded half-yearly.
Answer:
Let the principal amount be P=Rs 6000.Rate of interest = 10% p.a.= 5% for half yearlyTime (n) = 1 year = 2 half yearsNow, A = P ×1+R100n= Rs 6000 ×1+ 51002 = Rs 6000 ×1051002= Rs 6000 ×2120× 2120= Rs 15×21×21= Rs 6615∴ CI = A-P = Rs (6615-6000) = Rs 615Hence, the amount is Rs 6615 and the CI is Rs 615.
Page No 157:
Question 4:
A sum amounts to Rs 23762 in 2 years at 9% per annum, compounded annually. Find the sum.
Answer:
Amount (A) = Rs 23762Rate of interest (R) =9%Time (n) = 2 yearsNow, A = P ×1+R100n⇒Rs 23762 = P ×1+91002 ⇒Rs 23762 = P ×109100×109100⇒P = Rs 23762×100×100109×109 ⇒P = Rs 20000∴ The principal amount is Rs 20000.
Page No 157:
Question 5:
A scooter is bought for Rs 32000. Its value depreciates at 10% per annum. What will be its value after 2 years?
Answer:
Let the principal amount be P=Rs 32000.Rate of interest (R) = 10%Time (n)= 2 yearsNow, A = Rs. P × 1-R100n = Rs. 32000×1-101002 = Rs. 32000×901002 = Rs. 32000×910×910 = Rs. 320 ×9×9 = Rs. 25920∴ The value of the scooter after 2 years is Rs 25920.
Page No 157:
Question 6:
Mark (✓) against the correct answer:
The compound interest on Rs 5000 at 10% per annum for 2 years is
(a) Rs 550
(b) Rs 1050
(c) Rs 950
(d) Rs 825
Answer:
(b) Rs. 1050
P= Rs. 5000R= 10%n= 2 yearsNow, A = Rs. P ×1+R100 n = Rs. 5000×1+101002= Rs. 5000×1101002 = Rs. 5000×1110×1110 = Rs. 50×11×11= Rs. 6050∴ CI = A-P = Rs. ( 6050-5000) = Rs. 1050
Page No 157:
Question 7:
Mark (✓) against the correct answer:
The annual rate of growth in population of a town is 5%. If its present population is 4000, what will be its population after 2 years?
(a) 4441
(b) 4400
(c) 4410
(d) 4800
Answer:
(c) 4410
Present population =4000
Rate of growth = 5%
To find the population of the town after 2 years, we have:
A = P ×1+R100n= 4000×1+51002= 4000×1051002 = 4000×2120×2120= 10×21×21 = 4410
Page No 157:
Question 8:
Mark (✓) against the correct answer:
At what rate per cent per annum will Rs 5000 amount to Rs 5832 in 2 years, compounded annually?
(a) 11%
(b) 10%
(c) 9%
(d) 8%
Answer:
(d) 8%
Here, A = Rs. P ×1+R100n⇒Rs. 5832= Rs. 5000×1+R1002⇒Rs. 5832 Rs. 5000= 1+R1002 ⇒27252 = 1+R1002⇒1+R100= 2725⇒R100 = 2725-1 =27-2525=225∴R = 100×225= 8%
Page No 157:
Question 9:
Mark (✓) against the correct answer:
If the simple interest on a sum of money at 10% per annum for 3 years is Rs 1500, then the compound interest on the same sum at the same rate for the same period is
(a) Rs 1655
(b) Rs 1155
(c) Rs 1555
(d) Rs 1855
Answer:
(a) Rs. 1655
Here, SI =P×R×T100⇒Rs. 1500 = P× 10×3100⇒P = 1500×10010×3= Rs. 5000Now, A = P ×1+R100n= Rs. 5000×1+101003 = Rs. 5000× 1101003 = Rs. 5000×1110×1110×1110 = Rs. 5×11×11×11 = Rs. 6655∴ CI = A-P =Rs (6655-5000) = Rs 1655
Page No 157:
Question 10:
Mark (✓) against the correct answer:
If the compound interest on a certain sum for 2 years at 10% per annum is Rs 1050, the sum is
(a) Rs
3000
(b) Rs 4000
(c) Rs 5000
(d) Rs 6000
Answer:
(c) Rs. 5000
Here, A = P×1+R100n = P×1+101002 = P×1101002 = P× 1110×1110Now, CI= A-P⇒Rs. 1050 = 121p100- P = 121P-100P100= 21P100∴ P = Rs 1050× 10021 =Rs 5000
Page No 157:
Question 11:
Fill in the blanks:
(i) A=P1+.......
.100n.
(ii) (Amount) - (Principal) = .........
(iii) If the value of a machine is Rs P and it depreciates at R% per annum, then its value after 2 years is .........
(iv) If the population P of a town increases at R% per annum, then its population after 5 years is .........
Answer:
(i) A= P1+R100n(ii) Compound interest(iii) A= P1-R1002, where A is the value of the machine after 2 years(iv) A= P1+R1005, where A is the population of the town after 5 years
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